# Calculating cup products using cellular cohomology

Most algebraic topology books (for instance, Hatcher) contain a recipe for computing cup products in singular or simplicial homology. In other words, given two explicit singular or simplicial cocycles, they contain a recipe for computing an explicit cocycle representing the cup product of the cocycles in question.

Is there a similar recipe in cellular cohomology? In other words, if I have a very explicit CW complex and two explicit cellular cocycles, then is there a recipe for computing a cellular cocycle representing their cup product?

Of course, one answer is to subdivide everything up into a simplicial complex, but that is messy (and not always possible). Is there a better way?

I'm especially interested in the special case of 2-dimensional CW complexes, where the only interesting cup products are between elements of $H^1$.

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My favorite way to compute cup products is to use the serre spectral sequence. You can do the computation of products in $BS^1$ and $BC_2$ pretty easily. This does not address your particular complex since neither of these are finite dimensional, but you can restrict. I think this is like what charles is saying, except I am using Serre SS to do the group cohomoloy calculation. – Sean Tilson Oct 20 '10 at 23:32

It's easy to compute the cross product (i.e., the Kunneth map) $H^*X\otimes H^*Y\to H^*(X\times Y)$ in cellular homology, by means of the isomorphism $C_*X\otimes C_*Y\approx C_*(X\times Y)$ of complexes of cellular chains.

The obstruction to obtaining a formula for the cup product is the fact that the diagonal embedding $d:X\to X\times X$ is not a cellular map. The cellular approximation theorem tells you that $d$ is homotopic to some cellular map $f:X\to X\times X$, so cup product on cellular cochains is given by $C^*(X)\otimes C^*(X)\approx C^*(X\times X) \xrightarrow{f^\#} C^*(X)$.

I'm unaware of any general "formula" to find $f$ (you'd have to construct it inductively, one dimension at a time). But with only 2-dimensions, this shouldn't be at all bad (in any case, the calculation of the cup product $H^1\times H^1\to H^2$ mainly only depends on the fundamental group of $X$, since the map $X\to K(\pi_1X,1)$ inducing an isomorphism on $\pi_1$ also induces an isomorphism in $H^1$, .)

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Is computing cup products in group cohomology any easier than for spaces? Actually, the 2-complex I'm working with is already aspherical, so that would do it. – DanT Oct 20 '10 at 21:39
Maybe. My point is that your calculation only depends on the group cohomology calculation, which suggests that (however you carry it out), you should only need to worry about things related to the fundamental group. Since it is aspherical, this is certainly the case. – Charles Rezk Oct 20 '10 at 21:46
So suppose you have your space $X$ and call $G=\pi_1(X)$. Then you can do the group cohomology computation and if you have cocycles you can compute the yoneda product which wont be too bad (theoretically). The map that charles mentions that induces an iso on $H^1$ is a ring map since it comes from a map of spaces. I think that will work. – Sean Tilson Oct 20 '10 at 23:31

Cup products $\cup: H^1(X,\mathbb{Z})\times H^1(X,\mathbb{Z}) \to H^2(X,\mathbb{Z})$ are easy to visualize. Given a cellular 2-complex $X$, the 1-skeleton of $X$ is a graph $\Gamma$, and the 2-skeleton is a collection of disks with boundaries attached to the graph $\Gamma$, which (up to homotopy) are encoded by closed paths in $\Gamma$. The 1-cycles $\alpha_1,\alpha_2 \in H^1(X,\mathbb{Z})$ are encoded by maps $\alpha_i: X\to S^1=K(\mathbb{Z},1)$, and the 2-cycle $\alpha_1\cup\alpha_2$ is represented by the map $(\alpha_1\times\alpha_2)^{\ast}: H^2(S^1\times S^1)\to H^2(X)$. Choose the standard cell structure on $S^1 =c_0\cup c_1$. Subdivide $\Gamma$ by putting a vertex in the middle of each edge $e\subset \Gamma$ into intervals $e_1\cup e_2$. Then up to homotopy, we may assume that $\alpha_i$ sends $e_i$ to $c_0$ (and is extended over the 2-cells in any fashion). Then $\alpha_1\times \alpha_2: \Gamma \to (c_0\times c_1\cup c_1\times c_0)= S^1 \vee S^1\subset S^1\times S^1$, in such a way that each edge $e_1$ is sent to a horizontal edge, and each edge $e_2$ is sent to a vertical edge. To figure out the degree of this map, for each 2-cell $f\subset X$, we lift $\alpha_1\times \alpha_2:\partial f\to \mathbb{R}^2$, the universal cover of $S^1\times S^1$. Then we compute the winding number of the path represented by $f$ with respect to the midpoints of the lattice $(\widetilde{S^1\vee S^1} \subset \mathbb{R}^2$ giving $\alpha_1\cup\alpha_2(f)$. For example, the closed path in this picture has winding number 7.

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Cup products in a 2-dimensional CW complex with a single 0-cell should be computable directly from the definitions, once one knows how the 2-cells attach to the 1-cells. There is a discussion of this in the first chapter of Roger Fenn's book "Techniques of Geometric Topology" (Cambridge, 1983). There are also a few examples worked out in my algebraic topology book near the beginning of section 3.2, where the idea is to subdivide to get a Delta-complex (mild generalization of a simplicial complex) by adding a new vertex in the center of each 2-cell, and then one can use the simplicial cup product. The discussion in Fenn's book is more general than this, but he doesn't work out in detail the most general case.

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